Let $\phi$ a homomorphism from $G$ to $H$. Let $K \leq G$ contain the kernel of $\phi$. Show that $\phi ^{-1}\left( \phi\left( K \right) \right) = K.$
I have been asked to prove this statement. It looks quite simple to prove, but for some reason I am very stuck trying to prove it. I have taken the approach of trying to show that the sets on each side of the equality are subsets of each other. I have that the set on the left side of the equality is essentially $$ \left\{ g \in G: \phi\left( g \right)=\phi \left( k \right) \text{ for some }k\in K \right\} $$ I also see that since the kernel of $\phi$ is in $K$, then we have that $$ \phi\left( g \right)=e \implies g\in K $$ I am using the approach of trying to prove that if $ \phi\left( g \right)=\phi \left( k \right) \text{ for some }k\in K $, then $ g \in K $. Obviously the other containment $\left(( K \subseteq \phi ^{-1}\left( \phi\left( K \right) \right) \right)$ is quite trivial, but I cannot see is how to prove the harder containment. I know this is probably very easy, but for some reason I can't work it out. Thanks for any help.
Hint: $\phi$ is a homomorphism, so if $\phi(g)=\phi(k)$, then $\phi(gk^{-1})=?$